Answer
Please see the work below.
Work Step by Step
We know that
$Q_{water}=-Q_{copper}$
$m_{water} c_{water}\Delta T_{water}=-m_{copper}c_{copper}\Delta T_{copper}$
We plug in the known values to obtain:
$(1.0)(4184)(25-20)=(m_{copper})(386)(300-25)$
$m_{copper}=0.20Kg$
